Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, qu, f[100005], v[100005 + 100005], o[100005 + 100005], p, s[100005],
    r, fr[100005], to[100005], si[100005], curm, grp[100005], gr, rt[100005];
bool u[100005], vis[100005];
vector<int> b[100005], c[100005], q[100005], ans[100005];
void bd(int a, int b) {
  v[++p] = b;
  o[p] = f[a];
  f[a] = p;
}
void merge(vector<int> &a, vector<int> b) {
  vector<int> c;
  int l = 0, r = 0;
  while (l < a.size() || r < b.size()) {
    if (l == a.size())
      c.push_back(b[r++]);
    else if (r == b.size())
      c.push_back(a[l++]);
    else if (a[l] < b[r])
      c.push_back(a[l++]);
    else
      c.push_back(b[r++]);
  }
  a = c;
  if (a.size() > 11) a.resize(11);
}
void size(int x, int y) {
  s[x] = vis[x] = 1;
  for (int i = (f[x]); i; i = o[i])
    if (v[i] != y && !u[v[i]]) {
      size(v[i], x);
      s[x] += s[v[i]];
    }
}
void pick(int x, int y, int z) {
  int m = s[z] - s[x];
  for (int i = (f[x]); i; i = o[i])
    if (v[i] != y && !u[v[i]]) {
      pick(v[i], x, z);
      m = max(m, s[v[i]]);
    }
  if (m < curm) {
    curm = m;
    r = x;
  }
}
void group(int x, int y, int z) {
  grp[x] = z;
  for (int i = (f[x]); i; i = o[i])
    if (v[i] != y && !u[v[i]]) group(v[i], x, z);
}
void dfs(int x, int y) {
  c[x] = c[y];
  merge(c[x], b[x]);
  for (int i = (0); i < (q[x].size()); i++) {
    int d = q[x][i];
    if (vis[fr[d]] && vis[to[d]] && grp[fr[d]] != grp[to[d]]) {
      merge(ans[d], c[x]);
      rt[d] = r;
    }
  }
  for (int i = (f[x]); i; i = o[i])
    if (v[i] != y && !u[v[i]]) dfs(v[i], x);
}
void reset(int x, int y) {
  vis[x] = 0;
  for (int i = (f[x]); i; i = o[i])
    if (v[i] != y && !u[v[i]]) reset(v[i], x);
}
void root(int x) {
  size(x, 0);
  curm = 100005;
  pick(x, 0, x);
  grp[r] = 0;
  gr = 1;
  u[r] = 1;
  for (int i = (f[r]); i; i = o[i])
    if (!u[v[i]]) group(v[i], r, gr++);
  for (int i = (0); i < (q[r].size()); i++) {
    int d = q[r][i];
    if (vis[fr[d]] && vis[to[d]]) rt[d] = r;
  }
  for (int i = (f[r]); i; i = o[i])
    if (!u[v[i]]) dfs(v[i], 0);
  reset(r, 0);
  for (int i = (f[r]); i; i = o[i])
    if (!u[v[i]]) root(v[i]);
}
int main() {
  scanf("%d %d %d", &n, &m, &qu);
  for (int i = (1); i < (n); i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    bd(a, b);
    bd(b, a);
  }
  for (int i = (1); i <= (m); i++) {
    int t;
    scanf("%d", &t);
    b[t].push_back(i);
  }
  for (int i = (1); i <= (n); i++)
    if (b[i].size() > 11) b[i].resize(11);
  for (int i = (1); i <= (qu); i++) {
    scanf("%d %d %d", &fr[i], &to[i], &si[i]);
    q[fr[i]].push_back(i);
    q[to[i]].push_back(i);
  }
  root(1);
  for (int i = (1); i <= (qu); i++) {
    merge(ans[i], b[rt[i]]);
    if (ans[i].size() > si[i]) ans[i].resize(si[i]);
    printf("%d", ans[i].size());
    for (int j = (0); j < (ans[i].size()); j++) printf(" %d", ans[i][j]);
    puts("");
  }
  scanf("\n");
}